链表内元素的个数

Question

我想保存链表类对象实例内的元素数量。从下面的代码来看，每次我调用 addNodeFront() 或 addNodeBack() 函数时，成员变量 len 应该增加 1。但是，当我运行代码，getLen函数只返回1，而链表有2个元素。我应该从我的代码中修复什么？

#include <iostream>

class Node {
    public:
        int value;
        Node* next;
        
        Node(int value, Node* next) {
            this->value = value;
            this->next = next;
        }
};

class LinkedList {
public:
    Node* head;
    Node* tail;
    int len = 0;
    
    LinkedList() {
        this->head = nullptr;
        this->tail = nullptr;
    }
    
    LinkedList(Node* node) {
        this->head = node;
        this->tail = node;
    }

    int getLen() {
        return len;
    }
    
    void addNodeFront(Node* node) {
        if(head==nullptr && tail==nullptr) {
            this->head = node;
            this->tail = node;
            return;
        }
        Node* secondFirst = this->head;
        this->head = node;
        node->next = secondFirst;
        this->len++;
    }
    
    void addNodeBack(Node* node) {
        if(head==nullptr && tail==nullptr) {
            this->head = node;
            this->tail = node;
            return;
        }
        this->tail->next = node;
        this->tail = node;
        this->len++;
    }
    
    void addNodeAfterNode(Node* prevNode, Node* node) {
        if(prevNode == this->tail) {
            this->tail = node;
            prevNode->next = node;
            return;
        }
        node->next = prevNode->next;
        prevNode->next = node;
    }
    
    bool searchVal(int val) const {
        Node* s = this->head;
        while(s != nullptr) {
            if(s->value == val) return true;
            s = s->next;
        }
        return false;
    }
    
    void deleteNodeFront() {
        if(this->head==this->tail) {
            this->head = nullptr;
            this->tail = nullptr;
            return;
        }
        this->head = this->head->next;
    }

    void deleteNodeBack() {
        Node* secondLast = this->head;
        if(this->head==this->tail) {
            this->head = nullptr;
            this->tail = nullptr;
            return;
        }
        while(secondLast->next != this->tail) {
            secondLast = secondLast->next;
        }
        secondLast->next = nullptr;
        this->tail = secondLast;
    }

    void deleteNodeMiddle(Node* node) {
        if(node==this->head || node==this->tail) return;
        Node* prevNode = this->head;
        while(prevNode->next != node) {
            prevNode = prevNode->next;
        }
        prevNode->next = prevNode->next->next;
    }
    
    void traverseLinkedList() {
        Node* t = this->head;
        if(head==nullptr && tail==nullptr) std::cout << "Empty Linked List";
        while(t != nullptr) {
            std::cout << t->value << "->";
            t = t->next;
        }
        std::cout << "\n";
    }

};

int main() {
    Node node1(2,nullptr);
    Node node4(4,nullptr);
    LinkedList ls;
    ls.addNodeFront(&node1);
    ls.addNodeFront(&node4);
    std::cout << ls.getLen() << std::endl;
    ls.traverseLinkedList();
}

Answer 1

在下面的函数中：

LinkedList(Node* node) 
{
    this->head = node;
    this->tail = node;
}

只需添加以下行之一：

len++; //1
len = 1; //2

所以更新后的函数：

LinkedList(Node* node) 
{
    this->head = node;
    this->tail = node;

    len++; //1
    len = 1; //2
}

这是因为作为 head 节点 和 tail 节点 = node 现在，这意味着链表中有一个节点，所以我们应该给 len 加 1。